0
$\begingroup$

This is a statement in "Using Algebraic Geometry" by Cox et. al., Exercise 27 of Section 5.1:

Let $R=k[x_1,\dots,x_n]$. There exists unimodular rows $A$ with the property that no $\vec{f}\in R^l$ such that $A\vec{f}=1$ has an entry which is a nonzero element of $k$. In the case $R=k[x,y]$, the matrix $A=(1+xy+x^4, y^2+x-1, xy-1)$ is an example.

Background:

The goal of this Exercise is to investigate the syzygy of $A=(g_1,\dots,g_l)$, when $\langle g_1,\dots,g_l\rangle=1$. The previous parts have been devoted to showing that if $f_1g_1+\cdots+f_lg_l=1$ such that $f_i$ is a nonzero constant in $k$ for some $i$, then the syzygy form a free module. I have shown this.

My question:

I cannot see how the above matrix $A=(1+xy+x^4, y^2+x-1, xy-1)$ is an example in which when $\sum f_ig_i=1$, no $f_i$ can be a nonzero constant. I tried the following: $$f_1=a_0+a_1x+a_2y+a_3x^2+a_4xy+a_5y^2+\cdots\\ f_2=b_0+b_1x+b_2y+\cdots\\ f_3=c_0+c_1x+c_2y+\cdots\\ f_1g_1+f_2g_2+f_3g_3=0\implies\\ (a_0+a_1x+a_2y+\cdots)(1+xy+x^4)+(b_0+b_1x+b_2y+\cdots)(y^2+x-1)+(c_0+c_1x+c_2y+\cdots)(xy-1)=1 $$ Setting $f_1,f_2,$ or $f_3$ to be a constant, this then leads to a system of infinitely many equations. But how can I prove that this cannot stop in finite steps? Or is there a better way?

Thank you for your help!

  • 0
    In the title, you're making $f_i$ an element of $k$. In Cox, $f_i$ is a polynomial in the $x_j$. Please edit for consistency. – 2017-02-03
  • 0
    @GerryMyerson: I put $k[x,y]$ in the title and I am trying to show that $f_i$ cannot be in $k$. Isn't it what you meant or Cox meant? – 2017-02-03
  • 0
    The title says "no $f_i$ can be a nonzero element of $k$." The Cox quote has no $f_i$ in $R=k[x,y]$. In your display, $f_1=a_0+a_1x+\dots$ is clearly a member of $k[x,y]$. The title is inconsistent with the body. Please edit one or the other for consistency. – 2017-02-03
  • 0
    @GerryMyerson: I am sorry. I don't understand. $f_i$ is supposed to be in $R=k[x,y]$. So I set them that way, and trying to show that $f_1=a_0\ne 0$ leads to a contradiction. And similarly, $f_2=b_0\ne 0$, $f_3=c_0\ne 0$ both lead to contradiction. So the conclusion is no $f_i$ can be a nonzero element of $k$. I don't understand what I am missing. – 2017-02-04
  • 0
    You write, "Setting $a_0$, $b_0$, or $c_0$ to a constant...." But $a_0$, $b_0$, and $c_0$ *are* constants, whether you set them to be constants or not. It's $f_1$, $f_2$, $f_3$ that you're setting to be constants. – 2017-02-04
  • 0
    @GerryMyerson: I see! Thanks! – 2017-02-04

1 Answers 1

1

Assume$$1 = f_1 (1+xy+x^4) + f_2 (y^2 + x - 1) + f_3 (xy-1)$$ where $f_i \in k[x,y]$. It's easy to see that the set $\{1,1+xy+x^4,y^2+x-1,xy-1\}$ is linearly independent, therefore it is not possible that every $f_i \in k$.

Assume that $f_1$ is the only nonzero $f_i \in k$, then we must have $f_1 = 1$ for degree reasons. Now the $x^4$ term on the right side has to cancel with $f_2 x$, which shows that the $x-$degree of $f_2$ is $\geq 3$, but then $f_2 y^2$ contains a term with $(x,y)$-degree greater or equal to $(3,2)$, which can only cancel with $f_3 xy$. If that is the case, the degree in $x$ and $y$ of $f_3$ must be greater or equal to 1, which means that we are left with a term $xy$ (coming from $f_1 xy$) on the right hand side that is not cancelled out, so this case is not possible. The cases where $f_2, f_3$ are the only nonzero $f_i \in k$ can be dealt with similarly.

Now assume that there are two nonzero $f_i \in k$, let's say $f_1,f_2$. Then we have $f_1 - f_2 = 1$. We would need $f_1 xy$ to cancel either with $f_3xy$ or $-f_3$. In the first case, we would have $f_1,f_2,f_3 \in k$ which we already know is not possible. In the second case, we would have $f_3 = f_1 xy$ but then we are left with a $f_1 x^2y^2$ term that does not cancel out.

All the cases here can be handled this way, although it is not pretty.

  • 0
    Thank you for your answer! – 2017-02-03